内容分析讲稿

首都师范大学2005年数学分析解（10分求极

2n64.【解答

1 2nlim2 64 lim2 64lim11

n(31164)2 （12分设函

f(x)limx2n1ax2

x2n【解答

1,xx 1ab,x f(xax2bx,1x1f(x在1f(x1a ,x2 ,x 1a 1a故 limaxbx 1， limaxbx x1 a解得b1

x1 （12分已知函数f(x)在(ab上一致连续（ab为有限数。证明：f(x)在(ab上有界【解答f(x)在(ab上一致连续ab为有限数，故对任意0，存在(0,ba)，使得对任xy(ab)xy，就有f(xfy)xy(aa)xy(bbf(xfy)f(af(bf(a),xF(xf(xaxbF(x在[abF(x在(abff(b),x在(ab上有界（10分证明a0n1,2,，且limanl1，则lima0n【解答

naalimanl1，故对任意固定的1,lNN，使得对任意nN，恒有nan，a0，n1,2,，故 an，故对任意nN

0aan1an2aN10，n lima0n（12分f(x在[0,f"(x0f(0)0x0y0，f(xy)f(xfy.【解答函数f(x)在[0,上满足f"(x)0，故函数f(x)在[0,上连续且可导f(x在[0,f"(x0f'(x在[0,f'(2f'(1f(xy)f(max{x,y})min{x,

f(min{x,y})f(0)min{x,y}x0，y0f(0)0f(xyf(max{xyf(min{xyf(xfy（12分)))M0M2为正常数。证明：对任意x(0,，【解答

f'(x) xh(0,f(xhf(xf'(x)hf"(h2，于是2f'(x)f(xh)f(x)h

f"()h，因此2ff'(x) f"()hf(xh)f(x) h2M0M2ff(xh)f h都成立，2M0M2h的最小值为 x(0,)，有f'(x) （12分f(x在[0,f(x0

2M0M2h ，故对任意 xtf在(0,)内严格递增

(x)

x0f【解答f(x在[0,f(x0x(0, xxx'(x)[0tf(t)dt]'xf(x)0f(t)dtf(x)0tf(t)dtf(x)0(xt)fxxx

0 f(t)dt]' [0f(t)dt]2 [0f(t)dt]2故(x)在(0,内严格递增（12分f(x在[0,1上有二阶导1f(0)f(1)0，f"(x)0(x(0,1))，0f(x)dx01证明：(1)f(x在(0,1)(2)存在一点(0,1f'(【解答

f(t)dtf(x)在[0,1上有二阶导数，f"(x)0(x(0,1))，故f(x) 0f(x)dx0，故存在c(0,1)f(c0f(x)dx0f(xf(0)f(1)0f(00f(1)0f(x只有一个零点，则在(0,c和1上，f(x)0，故0f(x)dx0 ！故函数f(x)在(0,1)上恰好有两个零点1xF(x0f(t)dtF"(x)F(x)F"(x)F'(x)F'(x)F(x)[F'(x)F(x)]'[F'(x)F有零点，即证明ex[F'(x)F(x)]'[F'(x)F(x)]{ex[F'(x)F(x)]}'有零点，为此只要证明ex[F'(x)F(x有两个不同零点，即ex[F'(x)F(x)][exF(x)]'有两个不同)1G(x)xf(t)dt1f(t)dtxf(t)dt1f(t)dtxf(t)dt2xf

1

0f(t)dtxf(t)dtxf(t)dt2xf充分小邻域内G(x0，故G(x在(0,1)内有零点F(x在(0,1)内有零点F(x有（10分若级数an与cnanbncn，n证明级数bn【解答anbncn，故0cnbncnanan与cn都收敛，故正项级数(cnan收敛，故正项级数(cnbn)收敛，故级数bn也收敛。（12分I(x

1t2dt(1I(x在(0,limI(x0(2I(x)在(0,)【解答

(1xt(0,，有01t21t20I(x在(0,

1t2dt01t2dt limI(x0

0I(x)

01t

dt

dt 0，xx (2)

x1t2dtt

dta0xat0,01t

0 dt收敛 2在[1,)上单调递减有界 判别法 2dt在1

1I'(x01t2dt（12分证明函

(xy2)

x2y2f(x,y)

x2 x2y2在(0,0)点可微，但fx(x,y)在(0,0)点不连续f(x,f(x,y)fx2(x,y) 时

x2y2

x2x2x2

(x,y)(0,0)， (x,y

f(xyf(0,0)0f(xy在(0,0)x2fx(xy(x,y)

fy(x,y)0 fx(x,y)2xsin x2 x2 x2(x,y

2x 0，

x2 x2 x2 不存在，

f(xy不存在

(x,y在(x,y)(0,0) x2 x2

(x,y 点不连续（12分Sx2y2z2R2SIS【解答由Gauss公式和轮换对称性，

5 5I

x3dydz

xdx

x2dxdydz

y2z2)dxdydz x2y2z2 x2y2z2 x2y2z2（12分f(xy在单位圆域x2y21上有连续的偏导数，且在边界曲线上取值为零。记【解答

I

xfx(x,y)yfy(x,y)x2y2

dxdyxr作广义极坐标变换